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Below are the output characteristics of a common emitter n-p-n transistor: enter image description here

I need to know why collector current becomes independent of the collector-emitter voltage after some time? I don't exactly know the reason.. I tried a lot but failed... Web didn't helped me. Kindly help

Also, why the constancy appears at higher collector emitter voltage as the base current increases?

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    \$\begingroup\$ Welcome to the site. Please quickly realise that this is not a free design house, homework-answering service or an on-line technical encyclopedia, copied out to you on demand. People will help you take the next step if your question shows that you've done as much as you possibly could on your own - which your post doesn't, I'm afraid. Please revise your question showing your work and findings so far, in considerable detail. Or delete the question if Internet searches give you your answer anyway. Again, a warm welcome to the site. \$\endgroup\$ – TonyM Jan 23 '18 at 13:51
  • \$\begingroup\$ Seriously, i tried but i didn't get the right answer.. i'm not an engineer...im a school student reading in class 12... and this question i asked on PSE...migrated here... \$\endgroup\$ – Selena Jan 23 '18 at 13:53
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    \$\begingroup\$ Note for your example in the active region \$I_C = \beta I_B\$ where \$\beta=100\$ for this device. (ignoring the early voltage) \$\endgroup\$ – sstobbe Jan 23 '18 at 14:45
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    \$\begingroup\$ Having asked that question of a semiconductor guy, he explained the bipolar operation as "For each electron entering the base region, opposing charges chase after that electron in their seeking to combine and cancel the electron. The base region is intentionally made NARROW so almost all the emiter-injected carriers will miss the electron and end up in the collector region. Hence we have the names of Emitter, which emits charges that chase the electron, & of the Collector, which gathers up the charges that miss the electron. Beyond some minimum Collector voltage, behavior changes little. \$\endgroup\$ – analogsystemsrf Jan 23 '18 at 17:36
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    \$\begingroup\$ If I were you I'd simplify things by researching why the reverse saturation current of a diode is almost flat - it is very related to your question and might be an easier place to get answers than looking directly for BJT answers. After all, the collector base region is reverse biased in this region and behaves the same way. \$\endgroup\$ – Andy aka Jan 23 '18 at 18:34
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...independent of the collector-emitter voltage after some time?

Is the X-axis time? Nope, it is Vce!

In reality the curve is not that flat, Ic increases somewhat with increasing Vce.

The reason why Ic is somewhat independent of Vce is because the Base-Emitter junction determines the current and the Base-Collector junction has little influence on that as long as the base-collector voltage exceeds a certain value.

That current determined by the Base-Emitter junction is itself set by the current flowing into the base: Ib. Multiply Ib by the current amplification factor beta will give you the collector current Ic. This is only true in the flat part of the curves!

This is not a detailed description of how a Bipolar Transistor works in this respect, I'm ignoring some less important effects.

If you really want to understand the nitty-gritty then you should read a book about semiconductor devices.

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    \$\begingroup\$ So the best answer to this question is read a book about semiconductor devices. \$\endgroup\$ – dirac16 Jan 23 '18 at 14:17
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Saturation occurs and hence the there is something called breakdown voltage after that even if you try to increase the volume current is seldomly increased as the during breakdown voltage all the atoms in the depletion layer are breaking and after that the doped material (atoms) in emitter are also broken(electron release) and hence after certain value it is very hard to increase the current the increase negligiably occurs. HENCE THE MAXIMUM POSSIBLE CURRENT as output so furhermore now the number of electrons are enough in the collector and hence no more could contribute to further increase in output current.

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Um... that's a set of FET curves. At low gate voltage it becomes a resistor.

So, you're asking why BJTs transition from resistor-mode to constant-I-mode? They don't!

Here's what NPN curves look like:

enter image description here

Simple version is: at low Vce it stops "transisting," and acts like a conductor.

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